Divide the following complex numbers. $ \dfrac{9-2i}{4+i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${4-i}$ $ \dfrac{9-2i}{4+i} = \dfrac{9-2i}{4+i} \cdot \dfrac{{4-i}}{{4-i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(9-2i) \cdot (4-i)} {(4+i) \cdot (4-i)} = \dfrac{(9-2i) \cdot (4-i)} {4^2 - (1i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(9-2i) \cdot (4-i)} {(4)^2 - (1i)^2} = $ $ \dfrac{(9-2i) \cdot (4-i)} {16 + 1} = $ $ \dfrac{(9-2i) \cdot (4-i)} {17} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({9-2i}) \cdot ({4-i})} {17} = $ $ \dfrac{{9} \cdot {4} + {-2} \cdot {4 i} + {9} \cdot {-1 i} + {-2} \cdot {-1 i^2}} {17} $ Evaluate each product of two numbers. $ \dfrac{36 - 8i - 9i + 2 i^2} {17} $ Finally, simplify the fraction. $ \dfrac{36 - 8i - 9i - 2} {17} = \dfrac{34 - 17i} {17} = 2-i $